第110页 - 江苏版实验班提优训练数学八年级上下册答案

3

$\frac{7}{4}$

1

-2a

$解:∵5x-6≥0$

$∴x≥\frac {6}{5}$

$∴当x=\frac {6}{5}时，$

$原式有最小值0.$

$解:∵x^{2} 最小值是0，$

$∴当x=0时，$

$原式有最小值\sqrt {7} .$

$ \begin{aligned}解:原式&=\sqrt {(a+3)^{2} } \\ &=a+3 \\ \end{aligned}$

$ \begin{aligned} 解:原式&=\sqrt {(4-x)^{2} } -\sqrt {(x-1)^{2} } \\ &=4-x-(x-1) \\ &=4-x-x+1 \\ &=5-2x \\ \end{aligned}$

D

C

D

-x

x²+1

1-x

$ \begin{aligned}解:原式&=4-(-2)-2-6 \\ &=-2 \\ \end{aligned}$

$ \begin{aligned} 解:原式&=\sqrt {x^{2} +\frac {1}{x^{2} }+2} -\sqrt {x^{2} +\frac {1}{x^{2} }-2} \\ &=\sqrt {(x+\frac {1}{x})^{2} } -\sqrt {(x-\frac {1}{x})^{2} } \\ &=|x+\frac {1}{x}|-|x-\frac {1}{x}| \\ &=x+\frac {1}{x}+x-\frac {1}{x}& \\ &=2x \\ \end{aligned}$